Answer:
K.E = 0.0075 J
Explanation:
Given data:
Mass of the ball = 1.5 kg
radius, r = 50 cm = 0.5 m
Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec
Now,
the kinetic energy is given as:
K.E = ![K.E=\frac{1}{2}I\omega^2](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2)
where,
I is the moment of inertia = mr²
on substituting the values, we get
![K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes1.5%5Ctimes0.5%5E2%5Ctimes0.2%5E2)
or
K.E = 0.0075 J
Answer:
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Explanation:
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Answer:
no.
Explanation:
because the mass of an object never changes.
Answer:
![v_{max}=52.38\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D52.38%5Cfrac%7Bm%7D%7Bs%7D)
![v_{100}=33.81](https://tex.z-dn.net/?f=v_%7B100%7D%3D33.81)
Explanation:
the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:
![\sum{F}=0=F_d-W](https://tex.z-dn.net/?f=%5Csum%7BF%7D%3D0%3DF_d-W)
![F_d=W](https://tex.z-dn.net/?f=F_d%3DW)
![kv_{max}^2=m*g](https://tex.z-dn.net/?f=kv_%7Bmax%7D%5E2%3Dm%2Ag)
![v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D%5Csqrt%7B%5Cfrac%7Bm%2Ag%7D%7Bk%7D%7D%20%3D%5Csqrt%7B%5Cfrac%7B70%2A9.8%7D%7B0.25%7D%7D%3D52.38%5Cfrac%7Bm%7D%7Bs%7D)
To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:
![\sum{F}=ma=W-F_d](https://tex.z-dn.net/?f=%5Csum%7BF%7D%3Dma%3DW-F_d)
![ma=W-F_d](https://tex.z-dn.net/?f=ma%3DW-F_d)
![ma=mg-kv_{100}^2](https://tex.z-dn.net/?f=ma%3Dmg-kv_%7B100%7D%5E2)
(1)
consider the next equation of motion:
![a = \frac{(v_{x}-v_0)^2}{2x}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%28v_%7Bx%7D-v_0%29%5E2%7D%7B2x%7D)
If assuming initial velocity=0:
(2)
joining (1) and (2):
![\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7B100%7D%5E2%7D%7B2x%7D%3Dg-%5Cfrac%7Bkv_%7B100%7D%5E2%7D%7Bm%7D)
![\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7B100%7D%5E2%7D%7B2x%7D%2B%5Cfrac%7Bkv_%7B100%7D%5E2%7D%7Bm%7D%3Dg)
![v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g](https://tex.z-dn.net/?f=v_%7B100%7D%5E2%28%5Cfrac%7B1%7D%7B2x%7D%2B%5Cfrac%7Bk%7D%7Bm%7D%29%3Dg)
![v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}](https://tex.z-dn.net/?f=v_%7B100%7D%5E2%3D%5Cfrac%7Bg%7D%7B%28%5Cfrac%7B1%7D%7B2x%7D%2B%5Cfrac%7Bk%7D%7Bm%7D%29%7D)
(3)
![v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B%5Cfrac%7B9.8%7D%7B%28%5Cfrac%7B1%7D%7B2%2A100%7D%2B%5Cfrac%7B0.25%7D%7B70%7D%29%7D%7D)
![v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B%5Cfrac%7B9.8%7D%7B%28%5Cfrac%7B1%7D%7B200%7D%2B%5Cfrac%7B1%7D%7B280%7D%29%7D%7D)
![v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B%5Cfrac%7B9.8%7D%7B%28%5Cfrac%7B3%7D%7B350%7D%29%7D%7D)
![v_{100}=\sqrt{1,143.3}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B1%2C143.3%7D)
![v_{100}=33.81](https://tex.z-dn.net/?f=v_%7B100%7D%3D33.81)
To plot velocity as a function of distance, just plot equation (3).
To plot velocity as a function of time, you have to consider the next equation of motion:
![v = v_0 +at](https://tex.z-dn.net/?f=v%20%3D%20v_0%20%2Bat)
as stated before, the initial velocity is 0:
(4)
joining (1) and (4) and reducing you will get:
![\frac{kt}{m}v^2+v-gt=0](https://tex.z-dn.net/?f=%5Cfrac%7Bkt%7D%7Bm%7Dv%5E2%2Bv-gt%3D0)
solving for v:
![v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B%20%5Csqrt%7B1%2B%5Cfrac%7B4gk%7D%7Bm%7Dt%5E2%7D-1%7D%7B%5Cfrac%7B2kt%7D%7Bm%7D%20%7D)
Plots:
False ,it does not only associate with motion of object